Problem: You have found the following ages (in years) of all 5 porcupines at your local zoo: $ 2,\enspace 18,\enspace 13,\enspace 2,\enspace 15$ What is the average age of the porcupines at your zoo? What is the variance? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 5 porcupines at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{2 + 18 + 13 + 2 + 15}{{5}} = {10\text{ years old}} $ Find the squared deviations from the mean for each porcupine. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $2$ years $-8$ years $64$ years $^2$ $18$ years $8$ years $64$ years $^2$ $13$ years $3$ years $9$ years $^2$ $2$ years $-8$ years $64$ years $^2$ $15$ years $5$ years $25$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{64} + {64} + {9} + {64} + {25}} {{5}} $ $ {\sigma^2} = \dfrac{{226}}{{5}} = {45.2\text{ years}^2} $ The average porcupine at the zoo is 10 years old. The population variance is 45.2 years $^2$.